" If the velocity v of a particle varies as the square of its displacement "x" then the acceleration varies as "

The velocity v of moving object varies with displacement u as v=3x^(2)-2 ,then the acceleration at x=2m is

The velocity of any particle is related with its displacement As, x = sqrt(v+1) , Calculate acceleration at x = 5m .

The escape velocity of a particle of mass m varies as

A particle located at "x=0" at time "t=0" starts moving along the positive "x" - direction with a velocity "v" that varies as "v=alpha sqrt(x)".The displacement "(x)" of the particle varies with time as "(alpha" is constant)

If the displacement of a particle varies with time as sqrt x = t+ 3

If a point is moving in a line so that its velocity at time t is proportional to the square of the distance covered, then its acceleration at time t varies as

The velocity of a particle varies with its displacement as v=(sqrt(9-x^2))ms^-1 . Find the magnitude of the maximum acceleration of the particle.

Velocity of a particle varies with time as v=4t. Calculate the displacement of particle between t=2 to t=4 sec.